1
$\begingroup$

Suppose we have a Taylor expansion for a function $f$ with respect to t up to $M$-th order.

$$ \begin{equation} T_M = \sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k = f(x) + f'(x)\Delta t + \frac{1}{2}f''(x)\Delta t^2 + \cdots \end{equation} $$

What would be the general form of $T_M^2$ with respect to $\Delta t$?

$$ \begin{equation} T_M\times T_M = (\sum^M_{k=0}\frac{1}{k!}f^k(x)\Delta t^k)^2 = (???) + (???)\Delta t + (???)\Delta t^2 + \cdots \end{equation} $$

Perhaps Multinomial theorem can help?

$\endgroup$
1
  • 1
    $\begingroup$ It's just the general form for squaring a polynomial, but you can drop higher order terms if all you want is the $M$th order approximation. Note: the $1/k!$ is inside the summation. $\endgroup$ May 27 '19 at 13:51
2
$\begingroup$

Since we are squaring a sum we can utilize $$ \left( \sum_{k=1}^{n} a_k \right)^2 = \sum_{k=1}^{n} \sum_{j=1}^{n} a_ka_j $$ Applying this formula onto our problem yields

$$ \left(\sum_{k=0}^{M} \frac{1}{k!}f^k(x)\Delta t^k\right)^2 = \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} $$ Taking a closer look at the coefficients of $\Delta t^{k+j}$, we see that there are $k+j+1$ different combinations of $k$ and $j$. For example ($k+j=3$ cf. Jose Brox' answer), we have the combinations for: $$ (k,j) \in \{(0,1),(1,2),(2,1),(3,0)\}$$

Thus we can rewrite the above sum $$ \sum_{k=0}^{M}\sum_{j=0}^{M} \frac{1}{k!}\frac{1}{j!}f^k(x)f^j(x)\Delta t^{k+j} = \sum_{k=0}^{M}\left[ \sum_{j=0}^{k}\frac{1}{j!}\frac{1}{(k-j)!}f^j(x)f^{k-j}(x)\right]\Delta t^k$$ where the coefficient corresponding to the $k$-th exponent is $$\sum_{j=0}^{k}\frac{1}{j!(k-j)!}f^j(x)f^{k-j}(x) $$

$\endgroup$
2
  • $\begingroup$ In the second last equation, is that truncated to the order $M$? $\endgroup$ Jun 5 '19 at 10:32
  • $\begingroup$ My fault, I think it should be $k$ instead of $M$ in the second sum term. $\endgroup$
    – ulfgar
    Nov 5 '19 at 15:58
0
$\begingroup$

Like with usual polynomial multiplication, the term of degree $k$ comes from adding up all products of two monomials whose degrees sum to $k$. For example, for degree $3$ you have $(0,3)+(1,2)+(2,1)+(3,0)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.